Question
So recently I was thinking about this: How many scalars are available in $4$ dimensions in General Relativity (without being redundant)? For example, with metric we can construct the following scalar:
$$ g^{\mu \nu} g_{\mu \nu} = 4 $$ is the same as: $$ (g^{\mu \nu} \otimes g^{\rho \kappa}) \cdot (g_{\mu \nu} \otimes g_{\rho \kappa} ) = 16 $$
We also have scalars like curvature, torsion, inner product of the riemann tensor with itself, etc.
Motivation
My motivation for doing so is as follows: GR is currently through (rank $2$ symmetric) tensors formulated as:
$$ R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu} = \frac{8 \pi G}{c^4} T_{\mu \nu} $$
Hence any solution of the above automatically satisfies:
$$ (R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu}) (R^{\mu \nu} - \frac{1}{2} R g^{\mu \nu}) = \bigg(\frac{8 \pi G}{c^4}\bigg)^2 T_{\mu \nu} T^{\mu \nu} $$
But note the later equation is written purely in invariant observables. I was wondering if General Relativity could also be written purely in terms of observables? If not how many short are we? Can the remaining variables be expressed as something invariant and not a scalar (not sure if it would be a tensor either)?